Answer:
0.32M
Explanation:
<u>Step 1:</u> Balance the reaction
K2CO3 + Ba(NO3)2 ⇔ KNO3 + BaCO3
We have a 20 mL 0.2 M K2CO3 and a 30mL 0.4M Ba(NO3)2 solution
SinceK2CO3 is the limiting reactant, there will remain Ba(NO3)2 after it's consumed and produced KNO3 + BaCO3
<u>Step 2: </u>Calculate concentration
To find the concentration of the barium cation we use the following equation:
Concentration = moles of the <u>solute</u> / volumen of the <u>solution</u>
<u />
<u>[Ba2+] </u> = (20 * 10^-3 * 0.2M + 30 * 10^-3 * 0.4M) / ( 20 + 30mL) *10^-3
[Ba2+] = 0.32 M
The concentration of Barium ion in solution is 0.32 M
The answer is B. The complete equation is C6H12O6 + 6O2 -->6H2O + 6CO2 + energy. So we can know that A and C and D is right. For B, the reaction release energy so it is exothermic reaction.
Answer : The molarity of 
solution is, 0.352 M
Explanation :
First we have to calculate the moles of 
Molar mass of = 278.1 g/mol
Now we have to calculate the moles of 
The balanced chemical equation is:
From the balanced reaction we conclude that
As, 1 mole of produced from 1 mole of
So, 0.07041 mole of produced from 0.07041 mole of
Now we have to calculate the molarity of
Therefore, the molarity of solution is, 0.352 M
2CH3COOH +Na2CO3 ----> 2CH3COONa + H20 + CO2
know you can find what all products formed from his reaction
